Problem: You have found the following ages (in years) of all 4 tigers at your local zoo: $ 22,\enspace 14,\enspace 7,\enspace 3$ What is the average age of the tigers at your zoo? What is the variance? You may round your answers to the nearest tenth.
Answer: Because we have data for all 4 tigers at the zoo, we are able to calculate the population mean $({\mu})$ and population variance $({\sigma^2})$ To find the population mean , add up the values of all $4$ ages and divide by $4$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{4}} x_i}{{4}} $ $ {\mu} = \dfrac{22 + 14 + 7 + 3}{{4}} = {11.5\text{ years old}} $ Find the squared deviations from the mean for each tiger. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $22$ years $10.5$ years $110.25$ years $^2$ $14$ years $2.5$ years $6.25$ years $^2$ $7$ years $-4.5$ years $20.25$ years $^2$ $3$ years $-8.5$ years $72.25$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{110.25} + {6.25} + {20.25} + {72.25}} {{4}} $ $ {\sigma^2} = \dfrac{{209}}{{4}} = {52.25\text{ years}^2} $ The average tiger at the zoo is 11.5 years old. The population variance is 52.25 years $^2$.